package com.ljc;

import com.ljc.dto.ListNode;

/**
 * @author ljchen
 * @date 2022-08-22
 * You are given the heads of two sorted linked lists list1 and list2.
 * Merge the two lists in a one sorted list. The list should be made by splicing together the nodes of the first two lists.
 * Return the head of the merged linked list.
 *
 *  * Definition for singly-linked list.
 *  * public class ListNode {
 *  *     int val;
 *  *     ListNode next;
 *  *     ListNode() {}
 *  *     ListNode(int val) { this.val = val; }
 *  *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 *  * }
 */
public class E21MergeTwoLists {
    public static void main(String[] args) {
        ListNode l1 = new ListNode(1);
        l1.next = new ListNode(2);
        l1.next.next = new ListNode(4);

        ListNode l2 = new ListNode(1);
        l2.next = new ListNode(3);
        l2.next.next = new ListNode(4);

        ListNode l = mergeTwoLists(l1, l2);
        System.out.println(l.toString());

    }

    /**
     * 递归
     */
    public static ListNode mergeTwoLists(ListNode l1, ListNode l2) {
        if (l1 == null) {
            return l2;
        } else if (l2 == null) {
            return l1;
        } else if (l1.val < l2.val) {
            System.out.println("1");
            System.out.println("l1.val = " + l1.val);
            System.out.println("l2.val = " + l2.val);
            System.out.println();
            l1.next = mergeTwoLists(l1.next, l2);
            return l1;
        } else {
            System.out.println("2");
            System.out.println("l1.val = " + l1.val);
            System.out.println("l2.val = " + l2.val);
            System.out.println();
            l2.next = mergeTwoLists(l1, l2.next);
            return l2;
        }
    }


    /**
     * 迭代
     */
    public static ListNode mergeTwoLists2(ListNode l1, ListNode l2) {
        ListNode prehead = new ListNode(-1);

        ListNode prev = prehead;
        while (l1 != null && l2 != null) {
            if (l1.val <= l2.val) {
                prev.next = l1;
                l1 = l1.next;
            } else {
                prev.next = l2;
                l2 = l2.next;
            }
            prev = prev.next;
        }

        // 合并后 l1 和 l2 最多只有一个还未被合并完，我们直接将链表末尾指向未合并完的链表即可
        prev.next = l1 == null ? l2 : l1;

        return prehead.next;
    }

}
